π Example 280. πβ Example of an Uncoupled System.
Solve the uncoupled system:
\begin{align*}
\frac{dx}{dt} \amp = -x \quad \text{with} \ x(0) = 0.6\\
\frac{dy}{dt} \amp = -2y \quad \text{with} \ y(0) = 1.2
\end{align*}
Solution.
Each equation can be solved on its own:
\begin{equation*}
x(t) = 0.6 e^{-t}, \qquad y(t) = 1.2 e^{-2t}
\end{equation*}
Even though they evolve separately, we think of the combined solution as a pair:
\begin{equation*}
(x(t), y(t)) = \left(0.6 e^{-t}, 1.2 e^{-2t}\right).
\end{equation*}
You might imagine \(x(t)\) describing a cup of coffee cooling while \(y(t)\) describes a soda warming up β two unrelated processes happening side by side.